Find the square root of the following numbers using the Prime Factorisation Method: (i) 729 (ii) 400 (iii) 1764 (iv) 4096

Step 1: Find the square root of each number using the Prime Factorisation Method.

• (i) 729 Find the prime factors of 729:

  $$729 = 3 \times 243$$ $$243 = 3 \times 81$$ $$81 = 3 \times 27$$ $$27 = 3 \times 9$$ $$9 = 3 \times 3$$

  So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = (3 \times 3 \times 3)^2 = 27^2$.

  $$\sqrt{729} = \sqrt{3^6} = 3^{6/2} = 3^3 = \boxed{27}$$

• (ii) 400 Find the prime factors of 400:

  $$400 = 2 \times 200$$ $$200 = 2 \times 100$$ $$100 = 2 \times 50$$ $$50 = 2 \times 25$$ $$25 = 5 \times 5$$

  So, $400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 = (2 \times 2 \times 5)^2 = 20^2$.

  $$\sqrt{400} = \sqrt{2^4 \times 5^2} = 2^{4/2} \times 5^{2/2} = 2^2 \times 5^1 = 4 \times 5 = \boxed{20}$$

• (iii) 1764 Find the prime factors of 1764:

  $$1764 = 2 \times 882$$ $$882 = 2 \times 441$$ $$441 = 3 \times 147$$ $$147 = 3 \times 49$$ $$49 = 7 \times 7$$

  So, $1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 = (2 \times 3 \times 7)^2 = 42^2$.

  $$\sqrt{1764} = \sqrt{2^2 \times 3^2 \times 7^2} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = \boxed{42}$$

• (iv) 4096 Find the prime factors of 4096:

  $$4096 = 2 \times 2048$$ $$2048 = 2 \times 1024$$ $$1024 = 2 \times 512$$ $$512 = 2 \times 256$$ $$256 = 2 \times 128$$ $$128 = 2 \times 64$$ $$64 = 2 \times 32$$ $$32 = 2 \times 16$$ $$16 = 2 \times 8$$ $$8 = 2 \times 4$$ $$4 = 2 \times 2$$

  So, $4096 = 2^{12} = (2^6)^2 = 64^2$.

  $$\sqrt{4096} = \sqrt{2^{12}} = 2^{12/2} = 2^6 = \boxed{64}$$

• (v) 7744 Find the prime factors of 7744:

  $$7744 = 2 \times 3872$$ $$3872 = 2 \times 1936$$ $$1936 = 2 \times 968$$ $$968 = 2 \times 484$$ $$484 = 2 \times 242$$ $$242 = 2 \times 121$$ $$121 = 11 \times 11$$

  So, $7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11 = (2 \times 2 \times 2 \times 11)^2 = (8 \times 11)^2 = 88^2$.

  $$\sqrt{7744} = \sqrt{2^6 \times 11^2} = 2^{6/2} \times 11^{2/2} = 2^3 \times 11^1 = 8 \times 11 = \boxed{88}$$

• (vi) 9604 Find the prime factors of 9604:

  $$9604 = 2 \times 4802$$ $$4802 = 2 \times 2401$$ $$2401 = 7 \times 343$$ $$343 = 7 \times 49$$ $$49 = 7 \times 7$$

  So, $9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7 = (2 \times 7 \times 7)^2 = (2 \times 49)^2 = 98^2$.

  $$\sqrt{9604} = \sqrt{2^2 \times 7^4} = 2^{2/2} \times 7^{4/2} = 2^1 \times 7^2 = 2 \times 49 = \boxed{98}$$

• (vii) 5929 Find the prime factors of 5929:

  $$5929 = 7 \times 847$$ $$847 = 7 \times 121$$ $$121 = 11 \times 11$$

  So, $5929 = 7 \times 7 \times 11 \times 11 = (7 \times 11)^2 = 77^2$.

  $$\sqrt{5929} = \sqrt{7^2 \times 11^2} = 7^1 \times 11^1 = 7 \times 11 = \boxed{77}$$

• (viii) 9216 Find the prime factors of 9216:

  $$9216 = 2 \times 4608$$ $$4608 = 2 \times 2304$$ $$2304 = 2 \times 1152$$ $$1152 = 2 \times 576$$ $$576 = 2 \times 288$$ $$288 = 2 \times 144$$ $$144 = 2 \times 72$$ $$72 = 2 \times 36$$ $$36 = 2 \times 18$$ $$18 = 2 \times 9$$ $$9 = 3 \times 3$$

  So, $9216 = 2^{10} \times 3^2 = (2^5 \times 3)^2 = (32 \times 3)^2 = 96^2$.

  $$\sqrt{9216} = \sqrt{2^{10} \times 3^2} = 2^{10/2} \times 3^{2/2} = 2^5 \times 3^1 = 32 \times 3 = \boxed{96}$$

• (ix) 529 Find the prime factors of 529:

  $$529 = 23 \times 23$$

  So, $529 = 23^2$.

  $$\sqrt{529} = \sqrt{23^2} = \boxed{23}$$

• (x) 8100 Find the prime factors of 8100:

  $$8100 = 100 \times 81$$ $$100 = 10 \times 10 = (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2$$ $$81 = 9 \times 9 = (3 \times 3) \times (3 \times 3) = 3^4$$

  So, $8100 = 2^2 \times 3^4 \times 5^2 = (2 \times 3^2 \times 5)^2 = (2 \times 9 \times 5)^2 = (18 \times 5)^2 = 90^2$.

  $$\sqrt{8100} = \sqrt{2^2 \times 3^4 \times 5^2} = 2^{2/2} \times 3^{4/2} \times 5^{2/2} = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = \boxed{90}$$